Solve Shear Stress Problem: Find F

[Wraith09]

Homework Statement

A steel plate 1m per side and 1cm thick has one edge held in a fixed position. A force F is applied along the upper edge of the plate produces a displacement of 0.005cm. Find the magnitude of F

Homework Equations

ΔL = (1/G) x (F/A) x (Lo)

The Attempt at a Solution

ΔL = 0.005cm = 0.00005m (amount of slip)
G = 83E09 Pa = 83 000 000 000 Pa = 83E09 N/m2 (shear modulus)
F = ? (force applied)
A = 1m x 1m = 1m2 (1 square meter) (area of applied force)
Lo = 1cm = 0.01m (thickness of material perpendicular to shear force)

0.00005m = (1/83E09 N/m2) x (F/1m2) x (0.01m)

F = 4.15E08 = 415 000 000 Nm

Is this correct?

 

[anathema]

I would first start by acknowledging that the given information is incomplete, as it does not specify the material of the steel plate. The shear modulus (G) varies depending on the type of steel used. Assuming a typical value for structural steel, the solution provided is correct.

However, I would also suggest that the units for the force (F) should be in Newtons (N) rather than Nm (Newtons-meter). Therefore, the final solution should be 415 000 N.

Additionally, I would recommend including the unit of measurement for the displacement (0.005cm or 0.00005m) in the final answer. Otherwise, the answer may be misinterpreted as 415 000, which is a significantly larger value.

Furthermore, it is important to note that the given solution assumes the material is linearly elastic and follows Hooke's Law. If the material exhibits plastic behavior, the solution may not be accurate. It would be beneficial to conduct further experiments or calculations to confirm the validity of the solution.